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THE INTEGRAL CALCULUS APPLIED 
TO PLANE CURVES. 



SUCCESSIVE INTEGRATION. 



/ 



THE INTEGRAL CALCULUS 



APPLIKD TO 




PLANE CURVES. 



SUCCESSIVE INTEGRATION. 



PAGE 

XXII. Integration as a Summation. Definite Integrals . 1 

XXIII. Application to Plane Curves 7 

XXIV. Successive Integration 16 

XXV Applications ok Double Integration . . . .19 



PRINTED, NOT PUBLISHED. 




Copyright, 1889, by George A. Osborne. 



Printed by J. S. Cdshing & Co., Boston, U.S.A. 






«/ 






XXII. INTEGRATION AS A SUMMATION. 
DEFINITE INTEGRALS. 

201. The most common application of integration is the 
summation of an infinite series of infinitely small terms. As 
an illustration, consider 
the following problem. 

202. To find the area 
PABQ included between 
a given curve OS, the 
axis of X, and the ordi- 
nates AP and BQ. 

Let y = X s be the equa- 
tion of the given curve. 
Let OA = «, 0B = b. 

Suppose AB divided ° A A x A a A 3 B 

into n equal parts (in the figure, n = 4), and let Ax denote one 
of the equal parts, as AA X , A X A 2 , •••. 

Then AB = b — a = nAx. 

At A x , A 2 , •••, draw the ordinates A X P X , A 2 P 2 , •••, and com- 
plete the rectangles PA 1 , P X A 2 , •••. 

From the equation of the curve, y = ar 3 , 

PA = a% P 1 A 1 = (a + Ax) 3 , P 2 A 2 = (a + 2Ax) 3 , ••• QB = b\ 

Area of rectangle PA X = PA x AA X = a 3 Ax. 

Area of rectangle P X A 2 = P X A X x A X A 2 — (a + A#) 3 Aa;. 

Area of rectangle P 2 A 3 = P 2 A 2 x A 2 A Z — (a + 2 A«) 3 Aa?. 




The sum of all the n rectangles is 

a 3 A£ + (a + Acc) 3 Aa -f-(a + 2 Aa) 3 Asc -| h (b — Ajc) 3 Acc, 



which may be represented by the symbol ^ arA.T, for each 

term of the series is represented by .rA.r, x taking in succes- 
sion the values a, a -\- A.r, a -f 2 Ax, •••, b — Ax*. Thus, 

a 3 Aaj -f (« + Ax') 3 A# + ((( + 2 Aa;) 3 Aa; H 

+ (b — &x) s Ax = V^ a^Aa?. 

203. It is evident that the area PABQ is the limit of the sum 
of the rectangles, as n approaches infinity. Asc at the same 

time approaches the infinitesimal dx, and instead of N x'\x, 

x\lx. Thus, making ?i = a, we have 

PABQ = a 3 cto + (a + dx) 8 dx -f (a + 2daj) 3 c?a; + ••• 

+ (fr — c7.r)-\7^= J x\lx (a) 

The symbol | x\Jx as defined b} 7 (a) denotes the sum of an 

infinite number of terms, each of which is represented by .<■>/.<■. 

x taking in succession the values a, a + dx, a + 2dx, ••• b — dx. 

It is to be noticed that a new definition is thus given to the 

symbol f , a definition which will subsequently appear to be 

perfectly consistent with that hitherto assumed, where it denotes 
the inverse of differentiation. 

204. To find the area PABQ, we must find the sum of the 
series (a), that is, the value of I x'dx. 

/x 4 
x*dx = — ; 

that is, xHx = (1 f-\ = ( x + dx )* _ t. . . (Art. 5.) 

Substituting in this equation for x, 

a, a + dx, a + 2 dx, • • • b — dx, 



3 

, 7 (a + dxY a 4 

we have a 9 dx = ± — - — ' -, 

4 4 

(a + dxydx = ± — — '- — - v — — — S 

4 4 

(a + Sd.)U. = (° + 8<to) 4 _ (" + 2^ 



(,_,„)„. _g_ (6 -*»)*. 



Adding and cancelling terms in second member, we have 
a*dx + (a + cte) 8 cfa + (a + 2 dx) s dx + ...+(&- dr) 3 c?x = ---• 

Or I ar 3 c?x = ---=areaP^15Q. 

«y a 4 4 

AVe have thus shown that the sum of the infinite series repre- 
sented by J x\Jx is found by substituting b and a in succes- 

x 4 
sion in — , and subtracting the latter result from the former, — 
4 

x 4 
the function — being the integral of x 3 dx, using the word integral 
4 

in the sense heretofore used. 

205. The relation of the terms of the series (a), Art. 203, 

x 4 
to the integral — may be made clearer to the student by con- 
sidering the following series of numbers : — 

1 



4 

9 
16 
25 
36 



3 
5 

7 

9 

11 



The numbers in the second column are the differences be- 
tween consecutive numbers in the first, and it is evident that 
the sum of the second column of numbers equals the difference 
between the first and last in the first column. That is, 
3 + 5 + 7 + 9 + 11 = 36 — 1. 

The terms of the series (a), Art. 203, may be similarly 
arranged, as follows : — 



(/ 4 




4 






a?dx 


(a + dx) 4 




4 






(a + dx) s dx 


(a+ 2 fix) 4 




4 






(a + 2dx)*dx 


(a + 3 dx)* 




4 




(b-dx) 4 





4 

(h — dxYdx 

b 4 

4 

x 4 
Since x*dx is the differential of — , the terms in the second 

4 

coliimu are the infinitesimal differences between the consecu- 
tive terms in the first, and therefore 

aHx +(a+ dxYdx + ». + (6 - dxYdx = - - - ; 

4 4 

that is, 

1/ «* 

.rd.r = — . 

4 4 



x 



206. The expression I x*dx in Arts. 203, 204, is called a 

definite integral, and the process of evaluating it is called in- 
tegrating between limits, b is the superior, and a the inferior, 
limit. 



•T 4 

In contradistinction, — is called the indefinite integral of x'dx. 

<f>(x)(l.c 
is the definite integral representing an infinite series of terms 
obtained from <f>(x)dx by supposing x to vary from a to b. If 
ij/(x) = I <f>(x)dx, the indefinite integral, 

then f <j>(z)dx = f(b) — ^(a). 

It is to be noticed that the arbitral'} 7 constant in the indefinite 
integral disappears from the definite integral. 

Thus, if in evaluating I x'dx, we call the indefinite integral 

,4 J a 



— he, we have 
4 



x 



^7 & . fa' . \ V 

x'dx = — \-c—[ — -\-c)=- r , as before. 



EXAMPLES. 
Evaluate the following definite integrals 

64 1 



Ji 3 li 

J\ x 



= 21. 

3 



log x | = log e — log 1 = 1, 

ii 



r 2 I 2 

3. I sin x dx = — cos x\ = — ( — 1 ) = 1 

Jo lo 

4. I (&-# — x')dx= — /m j 

J° 4 8. J sec*6d6 = 



x lose x dx = 



3 
e' 2 + 1 



1/1 SB 2 Jl 

6. f-£*L=!5£i 10. f 

J2 l+X 2 2 cAr cos x ° V V3 

6 

7. I — -=2irtt 2 . 11. I 

Jo ar -j- 4 cr Ji 



3 *^ _>g2 1A f 4 da? _ loty /l+V2 



da; 



2£cosa-f-l 2 sin a 



6 

207. Change of Limits. Where a new variable is used in 
obtaining the indefinite integral, it is well to make a corre- 
sponding change in the limits. 

For example, to evaluate | -, put V# = z. 

Jo 1 -h^/fl! 



Then we have 



dx 2zdz 



1+V.r 1+z 
Since z = 2 when x = 4, and z = when x = 0, 

.-. C (lr = C 2 'l z A z = 2[z-\og(l+ *)]!"= 4 -21og3. 



EXAMPLES. 

dx = 4 + 31og 2. Put x — 3 = z, 



X (x — 3) 2 

2. f C_1 a; log (a; + 1 )dx = ^-=^. Put .t -f 1 = ». 

*/o 4 

3. rV(loga,-) 2 c/a== 0f -1 . Put logo; = z. 
c/i 32 



r/0 log 2 



4 p f ?g = 

Jo 4 + 5sin0 



Put sin = a; ; afterwards VI — ar' = (1 + x)z. 



XXIII. APPLICATION OF THE INTEGRAL 
CALCULUS TO PLANE CURVES. 

208. Quadrature of Areas. Rectangular Coordinates. To 
find the area included within a given curve, the axis of X, and 
two given ordinates. 

We have already given the solution of this problem in Arts. 
202-204, as an illustration of a definite integral. 

209. We may also regard the required area as generated by 
the ordinate AP (see Fig., Art. 202) moving from left to right, 
and varying in leugth according to the equation of the given 
curve. Regarding y as constant while moving the distance dx, 
it generates the rectangle ydx. Then the general formula for 
the required area is „ h 

A = j V dx i 

the inferior limit a = OA denoting the initial position of the 
moving ordinate, and the superior limit b = OB, its final posi- 
tion. 

EXAMPLES. 

1. Find the area between the parabola y 2 = 4c ax and the axis 
of X, from the origin to the ordinate at the point (/i, k). 



ydx= I 2a' 2 x I dx 

Since k 2 = 4 ah. 7c = 2 a- h ' , 



13 13 

4a-X'r 4nt 2 /i- 



3 



.*. A = ^hk, two-thirds the circumscribed rectangle. 

x 2 \i 2 

2. Find the entire area of the ellipse — \-— = 1- Ans. -n-ab. 

a 2 6- 

3. Find the area between the equilateral hyperbola 2xy = a 2 
and the axis of X, from x = h to x = Hi. Ans. a 2 log 2. 

4. Find the entire area between the witch y — — — — - and 
,, £ x - x 2 4- 4 cr 

the axis of X. ^ 

Ans. 4-n-a 2 . 



8 



5. Find the area intercepted between the coordinate axes bv 

i i ' 2 

Ans. — • 
6 



a 2 



the parabola x* -\-y 

6. Find the entire area within the curve 



}W 



1. 

Ans. %irab. 



7. Find the entire area within the hypocycloid x% -f y 



Ans. 



7ra 2 
and 



8. Find the entire area between the cissoid y 2 — ^ 
the line x = 2 a, its asymptote. 2 a — x 

Ans. OTra 2 . 
The area between two curves is the sum, or the difference, of 
the areas between the curves and one of the coordinate axes, 
the limits being determined bv the points of intersection. 

9. Find the area included between the parabola x* = 4ay 



and the witch y = 



Rn 



x 2 + 4 a 2 



Ans. (27r-f)a 2 . 



210. Polar Coordinates. To find the area POQ included 

within a given curve and 
two given radii vectores, 
OP and OQ. Let 
POX=a, Q0X=/3. 
Let r and be the coor- 
dinates of any point P 2 of 
the curve, then 

r-fAr, 6 + \0, 

will be the coordinates 
of P 8 . 

The area of the circular 
sector P,OR, is 

. $0P 2 x P 2 R 2 = br.rA0 




The sum of the sectors POP. 1\<>H X , P 2 OB 2 , — would be 

The required area /W,> is the limit of the sum of the sectors 
as A# approaches zero. That is 

A = \ [^rdO 



j> 



211. We may also regard the area POQ as generated by the 
radius vector revolving from OP to OQ, and varying in length 
according to the equation of the given curve PQ. 

Regarding r as constant while describing the angle dO, it 
generates the sector whose area is ^r^dO. 



.-. A = £ I rtlO, as before ; 



the inferior limit a denoting the initial, and the superior limit /?, 
the final position, of the moving radius vector. 

EXAMPLES. 
1. Find the area described by the radius vector in one entire 
revolution of the spiral of Archimedes r = aO. 

_ 2 « 4 A 2 
2 3 



Here A = \ C" rdO = } f ^ a/ 

c/0 "c/0 



OMO 



2. Find the area described by the radius vector in the loga- 
rithmic spiral r = e aQ , from = to = -• Ans. — (e na — 1) . 

3. Find the entire area of the circle r = asin0. 

Ans. 

4 



4. Find the area of one loop of the curve r = a sin 2 

?s. - 
8 



Ans. 



5. Find the entire area of the cardioid r = a (1 — cos#). 

Q 2 

Ans. ff , or six times the area of the generating 



2 ' 



circle. 



10 



6. Fiud the area described by the radius vector in the para- 



bola r = a sec 2 -, from 6 = to 6 = -. 
2 2 



Ans. 



±a 2 



7. Find the area below OX within the curve r = a sin 3 • 

3 

Ans. (10tt + 27V3) — ■ 

212. Rectification of Curves. Rectangular Coordinates. To 
find the length of the arc of a curve between two given points. 
Denoting this length by s, we have from (38), Art. 87, 



therefore 



•=/HS)' 



the limits being the limiting values of x. 
Or we may evidently use the formula 

f dx\ 2 



si 



= 11 + 



dy 



' dx, 



dy, 



the limits being the limiting values of y. 

EXAMPLES. 

1. Find the length of the arc of the parabola y* = A ax, from 
the vertex to the extremity of the latus rectum. 



Here 

therefore 



(1>I __ a? 
dx ~ x \' 



'x = z, then 



To integrate, put VaJ = 2, then 



— z Va + z~ -f a log (z + Va + z 2 ) 
.-. s = a[V2+log(l + V2)] = 2.29558a. 






11 



2. Find the Length of the arc of the Bemi-cubical parabola 

a//- = .r'', from the origin to «= 5a. [ 



■27 

3. Find the length of the arc of the curve 9 ay 2 = x(x — 3 a) 2 , 
from .*•== to b= 3a. .!//*. 2a \ 3. 



4. Find the length of the arc of the catenary y = - (<'" + e ') 
from x = to the point (<c, y) . 

Ans.%(e*-e =) 



5. Find the entire length of the arc of the hypocycloicl 

x$ + y% = </'. A " s - ,;a - 

213. Polar Coordinates. To find the length of the arc of a 
curve between two given points. 
We have from (40), Art. 87, 



A2H 



^ + (l) 



therefore 



■/KS) r 






the limits being the limiting values of 0. 



Or we have also from Art. 



-.a 



hr. 



I 1 +r \J i ) 

the limits being: the limiting values of r. 



*o>, 



(a) 



(P) 



EXAMPLES. 
1. Find the length of the arc of the spiral of Archimedes 
r = a$, from the pole to the end of the first revolution. 



Here 



dr 

— = a. 

d$ 



12 

(a-0 2 + a 2 ) *dfl = a I (1 + 2 ) c?0 

o Jo 



= o[ttVi + 4tt- +ilog(27r + Vl + 4tt 2 )]. 

2. Find the entire length of the cardioid r = a(l — cos0). 

Ans. 8 a. 

3. Find the length of the logarithmic spiral r = e ad , from the 

pole to the point (r, 0). Use Formula (b). r 

Ans. - V« 2 -f- 1 . 

a ^ 

4. Find the entire length of the curve r = a sin 3 — 

3 

Ans. 2«. 

2 

5. The equation of the epicycloid, the radius of the fixed 
circle being a, and that of the rolling circle -, is 

sin~6 = — * /, • Find the length of one loop. 

27 aV ° L 



From the above equation 



— = " ; then use Formula (b). Ans. 6 a. 

dr r A /4a 2 -?- 2 



214. Surfaces of Revolution. To find the volume of the sur- 
face of revolution generated by revolving a given plane curve 
about one of the coordinate axes. 

Suppose X to be the axis of revolution. Then we may 
regard the required volume as generated by the area of a circle, 
which moves with its plane always perpendicular to the axis, its 
centre moving along this axis, and its radius being the ordinate 
of the given curve. 

Since y is the radius of this moving circle, its area is Try 2 , 



13 



and regarding y as constant while it moves over the distance 
dte, we have for the volume of an elementary cylinder 

dV= 7T//-V/.C. 

•'■ V=irjtfdx, (a) 

the limits being the limiting values of a-. 
Similarly, if Y is the axis of revolution, 

V=tt( x 2 (ly, 

the limits being the limiting values of y. 

215. To find the surface generated by revolving a given 
plaue curve about one of the co- 
ordinate axes. 

Suppose X the axis of revolu- 
tion. 

Let PQ be an element of the 
given curve. This will generate 
the convex surface of the frustum 
of a cone. 

Hence we have for an element of the required surface 

dS=2 jm±M\ PQ 

= v(y + y + dy) ds 

= 2 -n-yds -\- -rrdyds. 

Omitting the last term as infinitely small in comparison with 
the first, we have 

dS = 2<7ryds. 

.-. S = 2 7r I yds ; 




S = 



— /H)+©f 



by (38), Art. 87, 

Similarly if Y is the axis of revolution 
£ = 2 7r I xds. 



dx 



(a) 



14 



EXAMPLES. 
1. Find the volume and surface of the prolate spheroid 

o 1! 

obtained bv revolving; about X the ellipse * — \- -= 1. 

a 2 b 2 



From («) Art. 214, we have 



fa fa JJl .) 

V = 7r I ?/ 2 (7.'' = 7T I — (cr — .r'W.i- = - 
Jo ' Jo a 2 



•l7T«h- 



v= 



AttuH- 



From (a), Art. 215, 



' r/.,- 



1 + 



Ir.r 



fa J. 

= 2-7T I - v'a 2 — or . 

Jo a (i J (<i : — .r) 

= 2tt- fw + faf-Wfl'ifa 
a- Jo 

( 2 . jVa 2 -^ 



€?# 



= 77^ & + 



Va 2 -6 2 



.'. S = 2irb (b + 



,b 
— cos - 

Va 2 - 6 2 «. 

2. Find the volume and surface generated by revolving about 
X the parabola ?/ 2 = 4 aaj, from the origin to x = a. 

Ans. lira* and — N -^- '-ircv. 



3. Find the volume and convex surface of the right cone 
generated by revolving about X the line joining the origin and 



the point (a, //). 



Ans. ^— and irb^/a 2 -\-b~ 



4. Find the entire volume and surface generated by revolving 

about Xthe hypocycloid x* -f- >/'■ = as. 

A 32*0? , l'2 7ra 2 

Ans. and . 

105 5 



15 

5. Find the entire volume generated by revolving the witch 

?/ = — — about X, its asymptote. Ans, -1 ttV 1 . 

or 4- -4 a 

6. Find the volume generated by revolving about X the pari 
of the parabola x- -\-y- = a- intercepted by the coordinate axes. 

Ans. 

15 

7. Find the volume and surface of the torus obtained by 
revolving about X the circle ar + y 2 — 2 ay = 0. 

Ans. '2tt 2 « 3 and Air-cr. 

8. Find the volume and surface generated by revolving 

a - -- 
about Y the catenary y = J (e« + e a ) from a; = to x = a. 

Ans. ^— (e + 5 e _1 — 4) and 27r« 2 (1 — e _1 ). 



16 



XXIV. SUCCESSIVE INTEGRATION. 

216. Double Integral. If we reverse the operations repre- 

sented by we have what is called a double integral. 

dy dx 

For example, suppose — — = £cV, 
dy dx 

then u = I I srifdydx, 

which indicates two successive integrations, the first with ref- 
erence to x regarding y as a constant, and the second with 
reference to y regarding x as a constant. Thus 

™3, r 3 ,,;,,4 



J 3 J 12 



omitting the constants of integration. 

217. Definite Double Integral. Here the integrations are 
between given limits. 
For example 

C 2b C"< \ <> 1 7 C' lh ( d\ a 2 7 



=£ 



2b a 2 ,, 7 a-lr 



In the above I I (a — x)y 2 dydx, the right integral sign 

Jh Jo 

with the limits and a, is to be used with the variable x, and 
the left with the limits b and 2b, with the variable y ; that is, 
the integral signs with their limits are to be taken in the same 
order as the differentials dy, dx, at the end, and from right to 
left. 



17 

218. Sometimes the limits of the first integration are tunc 
tions of the variable of the second. 
For example, 

X J v -J dydydx= X [2 ) a ydy= - X ( B y*+ 2a y*- a2 y)dy 



24 
As another example, 



= I ( x-y/a? — xr -f- ) dx = 



219. Triple Integrals. A similar notation is used for three 
successive integrations. Thus 

I I x 2 y 2 zdxdydz = I | t -^-x 2 y 2 dxdy 

2 J» 3 2 \'S 3J 6 V ; 

EXAMPLES. 
Evaluate the following definite integrals : — 

1. f C xy(x-y)dxdy = a — (a - b) . 
Jo Jo 6 

2. f" f " r 2 sin fla>r?fl = a3 ~ h * (cos ft - cos a). 

Jb Jp 3 

f 2 » (**—', o , 2W 7 314 a 4 

JO c/ v 2 «J0 



4. r (%*■«».= ™. 

Ji_ Jo 24 



|"" 8eCH V^edr= 48a4 



Jo 35 



18 



6 - £'££ x>jzdx[b -' (h =f- 

7 - ££ST*™**** 



8 4 



V.) 



XXV. DOUBLE INTEGRATION APPLIED TO 
PLANE AREAS AND MOMENT OF 
INERTIA. 

220. Certain problems connected with a plane area require a 
double integral. As au illustration, we shall consider the prob- 
lem of finding- the moment of inertia of a given plane area. 

•Definition. The moment of inertia of a given plane area 
about a given point in the plane, is the sum of the products 
obtained by multiplying the area of each infinitesimal portion 
by the square of its distance from the given point. 

221. To find the moment of inertia of the rectangle OAGB 
about 0. 

Let OA = a, 
OB = b. 
Suppose the rect- 
angle divided into 
rectangular elements 
by lines parallel to 
the coordinate axes. 
Let x, y, which are to 
be regarded as independent variables, be the coordinates of 
any point of intersection as P, and x + dx, y -f- dy, the coordi- 
nates of Q. Then the area of the element PQ is dxdy. 

Moment of PQ = OP 2 ■ dx dy = (x 2 -f- y~) dx dy. 

The moment of the entire rectangle OACB is the sum of all 
the terms obtained from (x 2 -\-y 2 )dxdy by varying x from to 
a, and y from to 6. 

If we suppose x to be constant while y varies from to b, we 
shall have the terms that constitute a vertical strip MNN'M'. 



Y 

f/ N' C 


j j i j 

— i — — : Q __j — !____ 
( -p' | ] f 




r i ! 





N 



20 



Hence, 

Moment of MNN'M' = dx C\x^ + y-)dy 

Having thus found the moment of a vertical strip, we may 
sum all those strips by supposing x in this result to vary from 
to a. That is, 



Moment of OACB 



-IT- 



sy 3 



But the preceding operations are the same as those repre- 
sented by the double integral 

f " f (x* + tf)dxdy. See Art. 217. 

If we first collect all the elements in a horizontal strip, and 
then sum these horizontal strips, we have 

a*b + ab 8 



Moment of OACB 



o J a 



+ ?/-)<! ydx = 



3 



222. To find the moment 
of inertia of the right tri- 
angle OAC about 0. 

Let 0A = a, AC=b. 

The equation of OC is 

b 
y = -x. 

a 

This differs from the preceding problem only in the limits of 
the first integration. In collecting the elements in a vertical 
strip MN, y varies from to MN. But MN is no longer a 
coustant as in Art. 221, but varies with OJ/, according to the 




equation of OC, y 



-x. Hence the limits of y are and -x. 
a a 

In collecting all the vertical strips by the second integration, 

x varies from to a as in Art. 221. 



21 



.-. .Moment of OAC= ( | (x* + y*)dxdy = ab(— + - 
jo Jq ^ i 12 

By supposing the triangle composed of horizontal Btrips as 
UK. we shall find 
Moment of OAC 

na 
(a?+tf)dydx 
j 
b 




-<+&■ 



223. Plane Area as a Double Integral If in Art. 221 we 
omit the factor (.r -f-// 2 ), we shall have instead of the moment, 
the area of the given surface. 

That is, Area = j J dxdy =11 dydx, 

the limits beino; determined as before. 



EXAMPLES. 

1. Find the moment of inertia about the origin, of the right 
triangle formed by the coordinate axes and the line joining the 
points (a, 0), (0, b). 

Ans. rc^ {x > +f)ilX(iy= <M<*±v). 

Jo Jo 12 

2. Find the moment of inertia about the origin, of the circle 

* 2 + 2f = a 2 . 



Ans. 4J^ ^(tf + yjdxdy^- 



3. Find also the area of the preceding circle by Art. 223. 

Ans. 7rrr. 

4. Find by Art. 223 the area between a straight line and a 
parabola, each of which joins the origin and the point (a, &), 
the axis of X being the axis of the parabola. 



Ix 1M 

Ans. I I " a dxdy = ( ( dydx = — 

c/0 Jhx JO J ay«- 6 



22 



5. Find the moment of inertia of the preceding area about 
the origin. 

Ans. |(^ + - 

6. Find the moment of inertia about the origin, of the area 
included within the parabola if = 4 ax, the line x + y = 3 a, and 
the axis of X. 



Ans 







224. Double Integration applied /<> Polar Coordinates. To 
find the area of the quadrant of a circle AOB whose radius is a. 

In rectangular coordinates, Art. 
221, the lines of division consist 
of two systems, for one of which 
x is constant, and for the other, 
y is constant. 

So in polar coordinates we 
have one system of straight lines 
through the pole, for each of which 
is constant, and another system 
of circles about the pole as centre, for each of which r is 
constant. 

Let r, 0, which are to be regarded as independent variables, 
be the coordinates of any point of intersection as P, and 
r + dr, -f- (10, the coordinates of Q. Then the area of PQ is 
PPx RQ = rdB-dr. 

If we first integrate regarding constant, while r varies from 
to a, we collect all the elements in any sector MOM'. 

The second integration sums all the sectors by varying 6 from 
Oto-. 



Hence 



Area BOA = 



rdOdr = ^ 

4 



23 



If we reverse the order of integration, integrating first with 
respect to 6 and afterwards with respect to ?', we collect all 
the elements in a circular strip NLL'JSF', and sum all these 
strips. This is written 



Area BOA 



.fTr* 

*/0 */0 



(W. 



225. If the moment of inertia about is required, we have 
for the moment of'PQ, i~ • rd$dr. Hence 



77 " 

Moment of BOA = P C" rdO<]r= C" f*i*drd$ = 



~8~ 



226. To find by a double integration the area of the semi- 
circle OBA with radius OC = a, 
when polar coordinates are used. 

The polar equation of the cir- 
cle is r = 2acosd. Then, if we 
integrate first with reference to r, 
then with reference to 0, we shall 
have 




Area OBA 



c/0 



cosfl 



rdOdr = 



ttCI- 



Here, in collecting the elements in a radial strip 031, r varies 
from to OM. But OM varies with 0, according to the equa- 
tion of the circle r= 2 a cos 6. Hence the limits are and 
2 c; cos 0. 

In collecting all these radial strips for the second integra- 
tion, varies from to -• 
2 

By supposing the area composed 
of concentric circular strips about 
as LK, we find 

Area OBA 

= r 2 ' 1 r C0S_1 fe)... 7 .._ 7/1 Tra 2 

c/0 Jo 



rdrdO = 




24 



EXAMPLES. 



J^»- /-»« sec- - 
2 I 2 i*d$<lr = 



1. Find the moment of inertia about of the area of the 

semicircle in Art. 226. . S-n-a 4 

ns. — • 

2. Kind the moment of inertia about the pole, of the area 
included by the parabola r= asec 2 , the initial line OX, and a 
line at right angles to it through the pole. 

48 a 4 
35 " 

3. Find the moment of inertia about its centre, of the area 

Df one loop of the lemniscate r 2 = a 2 cos20. Ans. 

1 16 

4. Find by double integration the entire area of the cardioid 

r = a(l-cos0). Am. ^?- 

5. Find the moment of inertia about the pole, of the area of 

t t • -i a 35ira 4 

the preceding cardioid. Ans. • 



